# Riemann curvature tensor derivation

1. Dec 23, 2004

### weio

Hey,

when calculating the Riemann curvature tensor, you need to calculate the commutator of some vector field $$V$$, ie like this :-

$$[\bigtriangledown_a, \bigtriangledown_b]$$ = $$\bigtriangledown_a\bigtriangledown_b - \bigtriangledown_b\bigtriangledown_a$$ = $$V;_a_b - V;_b_a$$

But why does this difference of antisymmtery give us the Riemman tensor?

thanks

2. Dec 24, 2004

### Rob Woodside

In addition to the commutator of the covariant derivatives, you need the commutator of the basis vectors too. Ignore torsion.

Think of your covariant derivative as a change along your basis. In a coordinate basis (defined by vanishing basis commutators) two different basis vectors span a plane. Think of a small quadrilateral spanned by the two basis vectors. When the vector V is carried around this quadrilateral your commutator of covariant derivatives gives the change in V. It's length can't change, but its direction does. So with R the full Riemann curvature:
[grada , gradb] V = R(.,V,a,b) . Riemann set up his geometry so it would look flat in the small. However, he was amazed that this difference resulting from taking a vector to nearby points could be described by an object (the full curvature tensor) that lived solely at the base point. This made him realize the importance of the curvature tensor and gave substance to his geometry.

3. Dec 24, 2004

### weio

Hey

So far this is how I understand it, though I know I could be very wrong. If you have two geodesics parallel to each other, with tangents $$V$$ and $$V'$$ , in which the coordinate $$x^\alpha$$ point along both geodesics. There is some connecting vector $$w^\alpha$$ between them. Let the affine parameter on the geodesics be $$\lambda$$

Riemman tensor calculates the acceleration between these two geodesics. so you calculate the acceleration at some point A, A' on each geodesic , and subtract them. this gives you an expression telling how the components of $$w^\alpha$$ change.

$$\frac{d^2w^\alpha} {d\lambda^2} = \frac{d^2x^\alpha} {d\lambda^2} | A' - \frac {d^2x^\alpha} {d\lambda^2} | A = - \Gamma^\alpha_0_0_\beta w^\beta$$

After that you calculate the full 2nd covariant derivative along V, ie , you get something like
$$\bigtriangledown v \bigtriangledown v w^\alpha = (\Gamma^\alpha_\beta_0_0 - \Gamma^\alpha_0_0,\beta) w^\beta$$
$$= R^a_0_0\beta w^\beta$$
$$= R^a_u_v_\beta V^u V^v w^\beta$$

That's where the tensor arises. so basically it's a difference in acceleration as geodesics don't maintain their seperation in curved space.

4. Dec 24, 2004

### Rob Woodside

Yes it arises there and in many other places, including the one you asked about and that I told you about.

5. Dec 27, 2004

### weio

Thanks!

I understand the derivation now. I found a simple one which fully explains it. for the curious, here is the link :- http://www.anasoft.co.uk/physics/gr/reimann/reimann.html [Broken]

weio

Last edited by a moderator: May 1, 2017